161ASpring2010-Lecture10

# 161ASpring2010-Lecture10 - Enthalpy H U + PV - the enthalpy...

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1 q p = H isochoric: isobaric: in both cases, q does not depend on the path from 1 to 2. H U + PV - the enthalpy q v = U Energy is often described as the capacity to do work. PV also has the capacity to do work. (compressed air can power an engine or a tool or a kid’s toy). Enthalpy includes the mechanical capacity of the system to do work. Why does is enthalpy close to energy at low T and diverges at higher T? (hint: think of the ideal gas law) Enthalpy

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2 Volume (V) P ex V final V initial Irreversible work A = p V Reversible Work A = RTln[Vf/Vi] P=nRT/V Reversible vs. Irreversible Change Isobaric expansion: w= = -p ex V Irreversible (why?) Reversible (isothermal) expansion: w=-nRTln{V f /V i } Which is greater? Consider indicator diagram » Work(rev.) = Area under curve » Work (irrev.) = Area under rectangle Work (rev.) > W (irrev.) Reversible work is the maximum which can be done » True of PV work » True of all work » Why can’t more work be done?

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4 The p-V curve for an Adiabatic process V p adiabat : pV = constant isotherm : pV = constant γ = ( α +1)/ α In an adiabatic process, p , V , and T all change monatomic gas γ =(5/2)/(3/2) = 5/3 = 1.67 diatomic gas γ =(7/2)/(5/2) = 7/5 = 1.4 Adiabats (adiabatic p-V curves) are steeper than isotherms Adiabatic and isothermal processes are reversible --
5 Adiabatic Process in an Ideal Gas adiabatic ( thermally isolated system) The amount of work needed to change the state of a thermally isolated system depends only on the initial and final states and not on the intermediate states. 0 2 1 = q 2 1 = W dU V P V 1 PV = Nk B T 1 PV = Nk B T 2 1 2 V 2 An adiabata is “steeper” than an isothermal: in an adiabatic process, the work flowing out of the gas comes at the expense of its thermal energy its temperature will decrease. - - = - = - = - - 1 1 1 2 1 1 1 1 2 1 1 1 1 1 ) , ( 2 1 2 1 γ V V V P dV V V P dV T V P W V V V V 5/3 1.67 (monatomic), 7/5 =1.4 (diatomic), 8/6 1.33 (polyatomic) The pressure at everypoint on the adiabata obeys the following equation:

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6 Homework Problem Imagine that we rapidly compress a sample of air whose initial pressure is 10 5 Pa and temperature is 22 0 C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature? γ 2 2 1 1 2 2 2 1 1 1 V P V P T Nk V P T Nk V P B B = = = 1 2 1 2 1 2 1 1 1 2 1 2 1 1 2 1 1 2 T T V V T T V P T Nk V V P V V P P B = = = = - - Rapid compression – approx. adiabatic , no time for the energy exchange with the environment due to thermal conductivity -poor approx. for a bike pump, works better for diesel engines -How does a diesel engine ignite?
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## This note was uploaded on 09/08/2010 for the course CHEM 160 at San Jose State University .

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161ASpring2010-Lecture10 - Enthalpy H U + PV - the enthalpy...

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